MOS substitution: Difference between revisions

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Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = (a+c)\mathbf{X}b\mathbf{m}(0).</math> This is the mode of <math>T</math> that has the most <math>\mathbf{X}</math> steps near the end. If <math>T</math> is [[primitive]], let <math>r</math> be the count of <math>\mathbf{X}</math> steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>a+c,</math> <math>r</math>-steps in the filling MOS <math>F = a\mathbf{L}c\mathbf{s}(k)</math> come in exactly 2 sizes, <math>i\mathbf{L}+j\mathbf{s}</math> and <math>(i-1)\mathbf{L}+(j+1)\mathbf{s}.</math> Since the detempering of the imperfect generator of <math>T</math> occurs only once in <math>S</math>, <math>S</math> admits a particularly elegant well-formed binary (using two distinct generators) [[generator sequence]] of length <math>q = \frac{a+c}{\gcd(a,c)},</math> the period of the filling MOS. The generator sequence corresponds to the circle of <math>r</math>-steps in the filling MOS. Letting <math>\mathsf{GS}(g_1, ..., g_{q})</math> be this generator sequence, <math>g_j</math> is either <math>p\mathbf{m} + i\mathbf{L} + j\mathbf{s}</math> or <math>p\mathbf{m} + (i-1)\mathbf{L} + (j+1)\mathbf{s},</math> according as the <math>j</math>-th <math>r</math>-step in the sequence of stacked <math>r</math>-steps on the chosen mode of <math>F</math> is <math>i\mathbf{L} + j\mathbf{s}</math> or <math>(i-1)\mathbf{L} + (j+1)\mathbf{s}.</math> (We could have chosen to use the mode of <math>T</math> on the other extreme of its generator arc instead, which corresponds to taking the circle of <math>(a+c - r)</math>-steps in <math>F</math> and is thus also valid.) The generator of the template MOS serves as the "guide generator" for this generator sequence.

~~<!--~~

=== If the template is a primitive MOS, and for some perfect generators <math>p_T, p_F, \ \left|p_T\right|_\mathbf{X} = \left|p_F\right|,</math> then MOS substitution yields almost parallelograms in the lattice ===

With the additional assumption that the number of X's in a perfect generator ''p''''T'' of the template MOS be a generator class of the filling MOS, the generator sequence yields ''q'' parallel chains ''C''1,

..., ''C''''q'' of the aggregate generator. The offset between ''C''''i'' and ''C''''i''+1 is equal to subst(''p''''T'', '''X''', ''p''''F''), where ''p''''T'' and ''p''''F'' are perfect generators (of appropriate lengths) of the template and filling MOSes, respectively. The aggregate generator is subst((''p''''T'')''r'', '''X''', ''F''''r''), where ''F'' is the filling MOS.

~~Hence~~ in the GS,

==== Caveat ====

* the ~~perfect generator~~ of the ~~filling~~ MOS ~~corresponds to advancing from~~ '~~'C''''i''~~ to ~~''C''''i''+1;~~

MOS substitution scales in general still have somewhat messy lattices. If you want an especially nice lattice (with two generators in the GS, and only one generator different in that GS) you should additionally want the number of slot letters in the template MOS's generator to be the generator interval class of the filling MOS.

* the ~~imperfect~~ generator of the filling MOS corresponds to looping back to ''C''1 but on the next note of ''C''1, so it and the ''q'' − 1 notes thereafter are advanced by 1 note from any predecessor notes in the chains.

~~Hence these particular MOS substitution scales satisfy a property that we call ''almost parallelogram''{{User:Inthar/Template:adhoc}}~~. An '''e'''-equivalent scale is ''almost a parallelogram'' if there exist non-negative integers ''m'', ''n'', 0 < ''a'' < ''n'', 0 < ''b'' < ''n'', a vector '''a''', and two linearly independent vectors '''v''' and '''w''' such that the set of notes in the scale as a subset of the lattice of '''e'''-equivalent pitches is

~~<math>\{\mathbf{a} + i\mathbf{v}\}_{i=a}^{n-1} \cup \{\mathbf{a} + i\mathbf{v} + j\mathbf{w}\}_{(i,j) \in [n]_0 \times [m-2]_1} \cup \{\mathbf{a} + i\mathbf{v} + (m-1)\mathbf{w}\}_{i=0}^{b}.</math>~~

In the above case, {{nowrap| ''n'' {{=}} ''q'' | '''v''' {{=}} subst(''p''''T'', '''X''', ''p''''F'') | and '''w''' {{=}} subst((''p''''T'')''r'', '''X''', ''F''''r'')}}.

~~The converse is false, as the scale in 5 letters [9/8 28/27 9/8 64/63 9/8 28/27 243/224 28/27 64/63 567/512 64/63] is almost a parallelogram.~~

~~-->~~

=== MOS substitution scales have block balance at most 2 ===

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 Consider the mode of the template MOS <math>T = T(\mathbf{m},\mathbf{X}) = (a+c)\mathbf{X}b\mathbf{m}(0).</math> This is the mode of <math>T</math> that has the most <math>\mathbf{X}</math> steps near the end. If <math>T</math> is [[primitive]], let <math>r</math> be the count of <math>\mathbf{X}</math> steps in a chosen (reduced) generator of <math>T.</math> Since <math>r</math> must be coprime to <math>a+c,</math> <math>r</math>-steps in the filling MOS <math>F = a\mathbf{L}c\mathbf{s}(k)</math> come in exactly 2 sizes, <math>i\mathbf{L}+j\mathbf{s}</math> and <math>(i-1)\mathbf{L}+(j+1)\mathbf{s}.</math> Since the detempering of the imperfect generator of <math>T</math> occurs only once in <math>S</math>, <math>S</math> admits a particularly elegant well-formed binary (using two distinct generators) [[generator sequence]] of length <math>q = \frac{a+c}{\gcd(a,c)},</math> the period of the filling MOS. The generator sequence corresponds to the circle of <math>r</math>-steps in the filling MOS. Letting <math>\mathsf{GS}(g_1, ..., g_{q})</math> be this generator sequence, <math>g_j</math> is either <math>p\mathbf{m} + i\mathbf{L} + j\mathbf{s}</math> or <math>p\mathbf{m} + (i-1)\mathbf{L} + (j+1)\mathbf{s},</math> according as the <math>j</math>-th <math>r</math>-step in the sequence of stacked <math>r</math>-steps on the chosen mode of <math>F</math> is <math>i\mathbf{L} + j\mathbf{s}</math> or <math>(i-1)\mathbf{L} + (j+1)\mathbf{s}.</math> (We could have chosen to use the mode of <math>T</math> on the other extreme of its generator arc instead, which corresponds to taking the circle of <math>(a+c - r)</math>-steps in <math>F</math> and is thus also valid.) The generator of the template MOS serves as the "guide generator" for this generator sequence.
-<!--
-=== If the template is a primitive MOS, and for some perfect generators <math>p_T, p_F, \ \left|p_T\right|_\mathbf{X} = \left|p_F\right|,</math> then MOS substitution yields almost parallelograms in the lattice ===
-With the additional assumption that the number of X's in a perfect generator ''p''<sub>''T''</sub> of the template MOS be a generator class of the filling MOS, the generator sequence yields ''q'' parallel chains ''C''<sub>1</sub>,
-..., ''C''<sub>''q''</sub> of the aggregate generator. The offset between ''C''<sub>''i''</sub> and ''C''<sub>''i''+1</sub> is equal to subst(''p''<sub>''T''</sub>, '''X''', ''p''<sub>''F''</sub>), where ''p''<sub>''T''</sub> and ''p''<sub>''F''</sub> are perfect generators (of appropriate lengths) of the template and filling MOSes, respectively. The aggregate generator is  subst((''p''<sub>''T''</sub>)<sup>''r''</sup>, '''X''', ''F''<sup>''r''</sup>), where ''F'' is the filling MOS.
-Hence in the GS,
+==== Caveat ====
-* the perfect generator of the filling MOS corresponds to advancing from ''C''<sub>''i''</sub> to ''C''<sub>''i''+1</sub>;
+MOS substitution scales in general still have somewhat messy lattices. If you want an especially nice lattice (with two generators in the GS, and only one generator different in that GS) you should additionally want the number of slot letters in the template MOS's generator to be the generator interval class of the filling MOS.
-* the imperfect generator of the filling MOS corresponds to looping back to ''C''<sub>1</sub> but on the next note of ''C''<sub>1</sub>, so it and the ''q'' &minus; 1 notes thereafter are advanced by 1 note from any predecessor notes in the chains.
-Hence these particular MOS substitution scales satisfy a property that we call ''almost parallelogram''{{User:Inthar/Template:adhoc}}. An '''e'''-equivalent scale is ''almost a parallelogram'' if there exist non-negative integers ''m'', ''n'', 0 < ''a'' < ''n'', 0 < ''b'' < ''n'', a vector '''a''', and two linearly independent vectors '''v''' and '''w''' such that the set of notes in the scale as a subset of the lattice of '''e'''-equivalent pitches is
-<math>\{\mathbf{a} + i\mathbf{v}\}_{i=a}^{n-1} \cup \{\mathbf{a}  + i\mathbf{v} + j\mathbf{w}\}_{(i,j) \in [n]_0 \times [m-2]_1} \cup \{\mathbf{a} + i\mathbf{v} + (m-1)\mathbf{w}\}_{i=0}^{b}.</math>
-In the above case, {{nowrap| ''n'' {{=}} ''q'' | '''v''' {{=}} subst(''p''<sub>''T''</sub>, '''X''', ''p''<sub>''F''</sub>) | and '''w''' {{=}} subst((''p''<sub>''T''</sub>)<sup>''r''</sup>, '''X''', ''F''<sup>''r''</sup>)}}.
-The converse is false, as the scale in 5 letters [9/8 28/27 9/8 64/63 9/8 28/27 243/224 28/27 64/63 567/512 64/63] is almost a parallelogram.
--->
 === MOS substitution scales have block balance at most 2 ===